Question

x+4÷x-4 + x-2÷x-3 = 19÷3​

Answer 1

Step-by-step explanation:

I hope this will help you

Answer 2

$Given \:\frac{x+4}{x-4} + \frac{x-2}{x-3} = \frac{19}{3}$

$\implies \frac{(x+4)(x-3)+(x-2)(x-4)}{(x-4)(x-3)} = \frac{19}{3}$

$\implies \frac{ x^{2}+x-12+x^{2}-6x+8}{x^{2}-7x+12} = \frac{19}{3}$

$\implies \frac{2x^{2}-5x-4}{x^{2}-7x+12} = \frac{19}{3}$

$\implies 3(2x^{2}-5x-4) =19(x^{2}-7x+12)$

$\implies 6x^{2} - 15x - 12 = 19x^{2} - 133x + 228$

$\implies 19x^{2} - 133x + 228 - 6x^{2} + 15x + 12 = 0$

$\implies 13x^{2} - 118x + 240 = 0$

/* Compare above equation with ax²+bx+c = 0 , we get */

$a = 13, b = -118 \:and \: c = 240$

$Discriminant (D) = b^{2} -4ac$

$= (-118)^{2} - 4 \times 13 \times 240$

$= 1444$

/* By Quadratic Formula */

$x = \frac{-b \pm \sqrt{D}}{2a}$

$= \frac{ -(-118) \pm \sqrt{1444}}{2\times 13}$

$= \frac{118\pm38}{26}$

$x = \frac{118+38}{26} \:or \: \frac{118-38}{26}$

$\implies x = \frac{156}{26} \: Or \: x = \frac{80}{26}$

$\implies x = 6 \:Or \: x = \frac{40}{13}$

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